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Classical Physics

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Microeconomics

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Particle Physics

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Quantum Field Theory

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Creation and Annihilation Operators
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Lagrangians in Quantum Field Theory
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Path Integral Formulation
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Relativistic Quantum Field Theory

Quantum Mechanics

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The Differential Operator
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The Essential Mathematics of Quantum Mechanics
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The Observer Effect
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The Qubit
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Time Dependent Perturbation Theory
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Time Evolution and Symmetry Operations
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Semiconductor Reliability

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The Weibull Distribution

Solid State Electronics

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Band Theory of Solids
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Fermi-Dirac Statistics
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Intrinsic and Extrinsic Semiconductors
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The MOSFET
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The P-N Junction

Special Relativity

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4-vectors
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Energy and Momentum, E = mc2
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Lorentz Invariance
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Statistical Mechanics

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Black Body Radiation
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Entropy and the Partition Function
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The Harmonic Oscillator
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The Ideal Gas

String Theory

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Bosonic Strings
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Extra Dimensions
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Introduction to String Theory
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Kaluza-Klein Compactification of Closed Strings
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Strings in Curved Spacetime
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Toroidal Compactification

Superconductivity

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BCS Theory
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Introduction to Superconductors
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Superconductivity (Lectures 1 - 10)
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Superconductivity (Lectures 11 - 20)

Supersymmetry (SUSY) and Grand Unified Theory (GUT)

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Chiral Superfields
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Generators of a Supergroup
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Grassmann Numbers
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Introduction to Supersymmetry
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The Gauge Hierarchy Problem

test

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test

The Standard Model

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Electroweak Unification (Glashow-Weinberg-Salam)
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Gauge Theories (Yang-Mills)
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Gravitational Force and the Planck Scale
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Introduction to the Standard Model
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Isospin, Hypercharge, Weak Isospin and Weak Hypercharge
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Quantum Flavordynamics and Quantum Chromodynamics
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Special Unitary Groups and the Standard Model - Part 1
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Special Unitary Groups and the Standard Model - Part 2
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Special Unitary Groups and the Standard Model - Part 3
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Standard Model Lagrangian
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The Higgs Mechanism
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The Nature of the Weak Interaction

Topology

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Units, Constants and Useful Formulas

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Constants
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Formulas
Last modified: January 18, 2020

Grassmann and Clifford Algebras ------------------------------- Grassmann (aka Exterior) Algebra -------------------------------- The fundamental operation in Exterior Algebra is the Wedge product. The wedge product is a completely antisymmetrized tensor product. To see this consider the tensor product of 2 vectors, ui and vj (≡ Tij). u ⊗ v = (1/2)(uv + vu) + (1/2)(uv - vu) = symmetric part + antisymmetric part Lets focus on the antisymmetric part. We can now define the wedge product as: u ∧ v = (1/2)(uv - vu) = (1/2)[u,v] At first sight this may look like the Lie algebra. However, in general, it is not. Unlike the Lie algebra, which lives in the tangent space of a manifold, the exterior algebra lives in a vector space that is separate from the original space of vectors, V, (hence exterior). This space is referrred to as Λp(V) and is called the pth EXTERIOR POWER. In other words the wedge product is an operation on a vector space where the result is a member of a new space product and not an element of the original vector space. Furthermore, the Lie algebra satisfies the Jacobi identity whereas the wedge product generally does not. Algebraic Properties -------------------- Associativity: (u ∧ v) ∧ w = u ∧ (v ∧ w) u ∧ u = 0 Consider: (u + v) ∧ (u + v) = u ∧ u + u ∧ v + v ∧ u + v ∧ v 0 = u ∧ v + v ∧ u Therefore, u ∧ v = -(v ∧ u) Thereby, confirming the completely antisymmetric properties for vectors. s ∧ t = t ∧ s = st and s ∧ s = s2 for scalars s and t. s ∧ u = u ∧ s = su for a scalar s and vector, u. p-vectors --------- A p-vector is a (p,0) tensor which is completely antisymmetric. p-vectors are constructed from wedge products. However, the dimension of Λp (i.e the # of basis elements) for each type of p-vector has to be taken into account. The dimension is given by the formula for combinations: - - | D | D! | p | = ---------- - - p!(D - p)! The wedge product of a p-vector and q-vector produces a (p + q) vector. However, there is a limitation on p, q and p + q for a given D. If (p + q) > D then one of the indeces would have to be duplicated violating the requirement of complete antisymmetry. For example, for D = 3 the wedge product of a 2-basis vector with a 2-basis vector produces: Tijkl = ei ∧ ej ∧ ek ∧ el However, there are only 3 basis vectors in D = 3 therefore el would have to be ei or j or k. Examples, for D = 3: 0-vector: f = f(x1,x2,x3) 1-vector: u = u1e1 + u2e2 + u3e3 2-vector: v = v12(e1 ∧ e2) + v23(e2 ∧ e3) + v13(e1 ∧ e3) 3-vector: w = w123(e1 ∧ e2 ∧ e3) What would these look like for D = 4? 0-vector: f = f(x1,x2,x3,x4) 1-vector: u = u1e1 + u2e2 + u3e3 + u4e4 2-vector: v = v12(e1 ∧ e2) + v13(e1 ∧ e3) + v14(e1 ∧ e4) + v23(e2 ∧ e3) + v24(e2 ∧ e4) + v34(e3 ∧ e4) 3-vector: w = w123(e1 ∧ e2 ∧ e3) + w124(e1 ∧ e2 ∧ e4) + w134(e1 ∧ e3 ∧ e4) + w234(e2 ∧ e3 ∧ e4) Note: The coefficients in each case are f(xμ). The Wedge Product of p-vectors ------------------------------ We use the example of the wedge product of a 1-vector and a 2-vector in D = 3 to illustrate the process. (u1e1 + u2e2 + u3e3) ∧ (v12(e1 ∧ e2) + v23(e2 ∧ e3) + v13(e1 ∧ e3)) = u1 ∧ v12 ∧ e1 | v = (u1v12e1 ∧ (e1 ∧ e2) + u1v23e1 ∧ (e2 ∧ e3) + u1v13e1 ∧ (e1 ∧ e3) + (u2v12e2 ∧ (e1 ∧ e2) + u2v23e2 ∧ (e2 ∧ e3) + u2v13e2 ∧ (e1 ∧ e3) + (u3v12e3 ∧ (e1 ∧ e2) + u3v23e3 ∧ (e2 ∧ e3) + u3v13e3 ∧ (e1 ∧ e3) = u1v23e1 ∧ (e2 ∧ e3) + u2v13e2 ∧ (e1 ∧ e3) + u3v12e3 ∧ (e1 ∧ e2) = u1v23e1 ∧ (e2 ∧ e3) - u2v13e1 ∧ (e2 ∧ e3) + u3v12e1 ∧ (e2 ∧ e3) = (u1v23 - u2v13 + u3v12)(e1 ∧ e2 ∧ e3) Which is a 3-vector. Now consider the product of two 2-vectors in D = 3: (u12(e1 ∧ e2) + u23(e2 ∧ e3) + u13(e1 ∧ e3)) ∧ (v12(e1 ∧ e2) + v23(e2 ∧ e3) + v13(e1 ∧ e3)) u12v12(e1 ∧ e2) ∧ (e1 ∧ e2) = u12v23(e1 ∧ e2) ∧ (e2 ∧ e3) = u12v13(e1 ∧ e2) ∧ (e1 ∧ e3) = 0 u23v12(e2 ∧ e3) ∧ (e1 ∧ e2) = u23v23(e2 ∧ e3) ∧ (e2 ∧ e3) = u23v13(e2 ∧ e3) ∧ (e1 ∧ e3) = 0 u13v12(e1 ∧ e3) ∧ (e1 ∧ e2) = u13v23(e1 ∧ e3) ∧ (e2 ∧ e3) = u13v13(e1 ∧ e3) ∧ (e1 ∧ e3) = 0 This demonstrates the closure property of the algebra. The Cross-Product ----------------- For D = 3, the cross product is the wedge product of a 1-vector with a 1-vector. Proof: u = aex + bey + cez (1-vector) v = eex + fey + gez (1-vector) Here D = 3 and the resulting exterior space is Λ2. u ∧ v = (bg - cf)(ey ∧ ez) + (ag - ce)(ex ∧ ez) + (af - be)(ex ∧ ey) Where the basis is {(ey ∧ ez),(ex ∧ ez),(ex ∧ ey)} The coefficients are the same as those in the usual definition of the cross product of vectors in three dimensions. However, the wedge product produces a bivector instead of a vector. The cross product is a special case where the wedge product is is the same the Lie bracket. We can prove this by verifying that the cross product satisfies the Jacobi identity: a x (b x c) + b x (c x a) + c x (a x b) = 0 Proof: Using the definition of the triple vector product, a x (b x c) = (a.c)b - (a.b)c, this becomes: (a.c)b - (a.b)c + (b.a)c - (b.c)a + (c.b)a - (c.a)b = 0 because the dot product is commutative. Areas in the Plane ------------------ The p-vectors are an algebraic construction used in geometry to study areas, volumes, and their higher dimensional analogues. Consider the following: For D = 2, the area is the wedge product of a 1-vector with a 1-vector. Proof: u = (x3 - x1)ex + (y3 - y1)ey v = (x2 - x1)ex + (y2 - y1)ey Here D = 2 and the resulting exterior space is Λ2. u ∧ v = (x3 - x1)(y2 - y1)(ex ∧ ey) + (x2 - x1)(y3 - y1)(ey ∧ ex) Now (ey ∧ ex) = -(ex ∧ ey). Therefore, u ∧ v = (x3 - x1)(y2 - y1)(ex ∧ ey) - (x2 - x1)(y3 - y1)(ex ∧ ey) = (x3y2 - x3y1 - x1y2 + x1y1 - x2y3 + x2y1 + x1y3 - x1y1)(ex ∧ ey) Let (x1,y1) -> (0,0) A = (x3y2 - x2y3)(ex ∧ ey) - - = det| x3 y3 |(ex ∧ ey) | x2 y2 | - - This is the standard determinant form for the area of a parallelogram. Differential Forms (aka p-Forms) -------------------------------- Differential forms are an approach to multivariable calculus that is independent of coordinates. They appear in the mathematics of differential geometry. A p-form (aka Differential form) is a (0,p) tensor which is completely antisymmetric. p-forms are duals to p-vectors. The wedge product of p-forms lives in the space Λp(V*). Since Λ(V*) is the dual of Λ(V), dim(Λ(V*)) = dim(Λ(V)). All of the properties and constructions discussed previously apply to differential forms: The Exterior Derivative ----------------------- p-forms can be both differentiated and integrated. The EXTERIOR DERIVATIVE, d allows us to differentiate a p-form to obtain a (p + 1) form. The exterior derivative is: (d)μ ≡ (∂/∂xμ) Note that: d(dφ) = 0 since partial derivatives commute. Forms can be expanded in terms of wedge products and the exterior derivative. For example, for D = 3: 0-form: γ = f(x1,x2,x3) 1-form: α = α1dx1 + α2dx2 + α3dx3 2-form: β = β12(dx1dx2) + β23(dx2dx3) + β13(dx1dx3) 3-form: δ = δ123(dx1dx2dx3) Where the coefficients are f(x1,x2,x3). In general: ω = Σfjdxj j dω = ΣΣ(∂fj/∂xi)dxidxj ij Example: In D = 3, γ = 3x2yz dγ = (∂(3x2)/∂x)dx + (∂(3x2)/∂y)dy + (∂(3x2)/∂z)dz + (∂(y)/∂x)dx + (∂(y)/∂y)dy + (∂(y)/∂z)dz + (∂(z)/∂x)dx + (∂(z)/∂y)dy + (∂(z)/∂z)dz = 6xdx + dy + dz = α dα = (∂(6x)/∂x)dx + (∂(6x)/∂y)dy + (∂(6x)/∂z)dz) ∧ dy + (∂(1)/∂x)dx + (∂(1)/∂y)dy + (∂(1)/∂z)dz) ∧ dx + (∂(1)/∂x)dx + (∂(1)/∂y)dy + (∂(1)/∂z)dz) ∧ dz = 6dx ∧ dy since dx ∧ dx = dy ∧ dy = dz ∧ dz = 0 Leibnitz (Product) Rule ----------------------- The exterior derivative of the product of a p-form and a q-form is given by: d(α ∧ β) = dα ∧ β + (-1)pα ∧ dβ The result will have dimension, D = p + q + 1. Proof: Consider a 1-form and a 2 form (p = 1, q = 2). The result will have D = 4. Therefore, both forms need to be specified in D = 4. For brevity we only consider the first term of each to illustrate the process. α = fdx1 + ... β = g(dx2dx3) + ... d(α ∧ β) = d(fg(dx1dx2dx3)) = (df.g + f.dg) ∧ dx1dx2dx3 = (df ∧ dx1) ∧ g(dx2dx3) + fdx1dg ∧ dx2dx3 After reordering the 2nd term to match the order of the 1st term we get: d(α ∧ β) = (df ∧ dx1) ∧ g(dx2dx3) - dg ∧ fdx1dx2dx3 Or, d(α ∧ β) = dα ∧ β + (-1)pα ∧ dβ Volume (aka Top) Form --------------------- When p = D the form is called the VOLUME FORM. It has a dimension of 1 and establishs an orientation of the exterior power (manifold). A coordinate system (on an oriented manifold) is oriented if: dx1dx2 ∧ ... ∧ dxD is positive. A transformation between oriented coordinate systems has positive Jacobian. Note that the exterior derivative of the top form is 0 since it would result in a p + 1 form that is greater than the dimension of the exterior power. Example: For D = 3 ω = dx1dx2dx3 Change of Coordinates --------------------- The derivative of a tensor transforms under a change of coordinates as: ∇μVν = ∂μVμ + ΓνμλVλ Where Γνμλ are the Christoffel symbols. Lets look at how the exterior derivative transforms. The definition of the wedge product allows us to write dx0 ∧ .... dxn-1 = (1/n!)εμ1 ... μn dxμ1 ∧ .... dxμn since both the wedge product and the Levi-Civita symbol are completely antisymmetric. Under a coordinate transformation, ε stays the same while the 1-forms change according to: Basis: dxμ' = (∂xμ'/∂xμ)dxμ or dxμ = (∂xμ/∂xμ')dxμ' Components: ωμ' = (∂xμ/∂xμ'μ' For convenience we work with the 2-form. Lets look at how (dx1dx2) part transforms. dx1dx2 = ((∂x1/∂x1')dx1' + (∂x1/∂x2')dx2') ((∂x2/∂x1')dx1' + (∂x2/∂x2')dx2') = (∂x1/∂x1')(∂x2/∂x1')(dx1'dx1') + (∂x1/∂x2')(∂x2/∂x2')(dx2'dx2') + (∂x1/∂x1')(∂x2/∂x2')(dx1'dx2') + (∂x1/∂x2')(∂x2/∂x1')(dx2'dx1') = (∂x1/∂x1')(∂x2/∂x2')(dx1'dx2') - (∂x1/∂x2')(∂x2/∂x1')(dx1'dx2') = (∂x1/∂x1')(∂x2/∂x2') - (∂x1/∂x2')(∂x2/∂x1')(dx1'dx2') = (∂x1/∂x1')(∂x2/∂x2') - (∂x1/∂x2')(∂x2/∂x1')(dx1'dx2') = det(∂xμ/∂xμ')(dx1'dx2') Therefore, the antisymmetric properties of the wedge product automatically takes care of the need to include the Christoffel symbols in the calculation. Since the Christoffel symbols involve the metric tensor and its derivatives, the implication of this is that differential forms allow us to work with manifolds without the help of the metric. In this sense they are coordinate independent or 'coordinate free'. Relation to Gradient, Curl and Divergence ----------------------------------------- Recall: For D = 3: 0-form: γ = f(xμ) 1-form: α = α1dx1 + α2dx2 + α3dx3 2-form: β = β12(dx1dx2) + β23(dx2dx3) + β13(dx1dx3) 3-form: δ = δ123(dx1dx2dx3) Differentiating the above p-forms gives: The gradient (∇f) ----------------- dγ = (∂f/∂xμ)dxμ (a 1-form) Example: f = x2 + z2 df = 2xdx + 2zdz The Curl (∇ x α) --------------- dα = dα1dx1 + dα2dx2 + dα3dx3 (a 2-form) Example: α = ydx + z2dy + 0dz dα = (dy ∧ dx) + 2z(dz ∧ dy) = -(dx ∧ dy) + 0(dx ∧ dz) - 2z(dy ∧ dz) = -2z(dy ∧ dz) + 0 -(dx ∧ dy) The Divergence (∇.β) -------------------- dβ = dβ12dx1dx2 + dβ23dx2dx3 + dβ13dx1dx3 (a 3-form) Example: β = x2(dy ∧ dz) + y2(dz ∧ dx) + z2(dx ∧ dy) dβ = 2x(dx ∧ dy ∧ dz) + 2y(dy ∧ dz ∧ dx) + 2z(dz ∧ dx ∧ dy) = (2x + 2y + 2z)(dx ∧ dy ∧ dz) Note that: ∇ x ∇f = 0 And, for a 1-form: ∇.(∇ x ) = 0 Integration ----------- In multivariable calculus we have: ∫∫f(x1,x2)dx1dx2 = ∫∫f(x1',x2')Jx1'dx2' Where J = det(∂xμ/∂xμ') is the determinant of the Jacobian of the transformation between the 2 coordinate systems that keeps the area elements equal. From before we had that: (dx1dx2) = det(∂xμ/∂xμ')(dx1'dx2') gμ'ν' = (∂xμ1/∂xμ1')(∂xμ2/∂xμ2')gμν det(gμ'ν') = (det(∂xμ/∂xμ'))2det(gμν) In Euclidean space det(gμν) = 1 Therefore, √(det(gμ'ν')) = det(∂xμ/∂xμ') Therefore, (dx1dx2) = √(det(gμ'ν'))(dx1'dx2') = √(|g'|)(dx1'dx2') (g' = det(gμ'ν')) If we make the identication: dxdy <-> √(|g'|)(dx1dx2) it is easy ro see that integration under a change of coordinates is properly understood as the integration of the volume (top) form! Generalized Stokes' Theorem --------------------------- M = manifold, ∂M = boundary of the manifold. ∫α = ∫dα ∂M M ∂M is of dimension p, α is of dimension p. M is of dimension p + 1, dα is of dimension p. b Fundamental Theorem of Calculus: ∫f = ∫df (p = 1) a ∴ f(b) - f(a) = ∫∇f dx C Stokes' Theorem: ∫F.dr = ∫∫∇ x F (p = 2) ∂S S Gauss' Divergence Theorem: ∫F.dr = ∫∫∫∇.F (p = 3) ∂V V Role of the Metric ------------------ If V has an inner product defined on it then an inner product is induced on V* and vice versa. Consider the following: v = Σvi∂/∂xi (tangent vector) i α = Σαjdxj (cotangent vector) j v.α = Σvi∂/∂xj.(Σαjdxj) i j The metric tensor has the effect of converting one of the input 1-vectors to a 1-form. Thus, if have 2 vectors, u and v, the 'tensor machine' produces: g(v,u) -> (vi∂/∂xi)(αjdxj) = viαk What we want is for the 1-form to 'eat' the 1-vector and produce the same result as the 1-vector 'eating' the original function. This can be expressed as: df(v) = v(f) ∈ ℝ f is a scalar function so df is a 1-form. We can now demonstrate this 'eating' process as follows: [(∂f/∂xi)dxi)](vjj) ---------- ^ | 1-form (i.e. f = x2 -> 2xdx) = vj(∂f/∂xi)dxi.∂j = vj(∂f/∂xiij = vi(∂f/∂xi) = vii(f) = v(f) The last term is the directional derivative. It is a scalar. It should not be confused with ordinary derivative which is interpreted as the gradient of the function at a point. The gradient of a scalar function is a vector. However, the two are related: ----------------------------------------------------- Digression: The directional derivative represents the instantaneous rate of change in f along a curve in the direction of the unit tangent vector. It is a scalar. The directional derivative, D, is related to the gradient of a scalar function as follows: Dvf = ∇f.v Example: - - v = | √2/2 | (unit tangent vector) | √2/2 | - - Dvf = ∇f.v f(x,y) = x2y - - - - Dvf.v = | 2xy |.| √2/2 |    | x2 | | √2/2 | - - - - At the point (-1,-1) we get: - - - - | 2 |.| √2/2 | = √2 + √2/2 = scalar | 1 | | √2/2 | - - - - The analogy between the DD and the ordinary derivative is: f'(a) = lim h->0(f(a + h) - f(a))/h Dv(a) = lim h->0(f(v + hi) - f(a))/h Note: The gradient of a vector is a rank 2 tensor. v = vii + v2j + v4k ∇v = (∂V/∂x)i + (∂V/∂y)j + (∂V/∂z)k = i(∂/∂x)(vii + v2j + v4k) + ... = ii(∂v1/∂x) + ij((∂v2/∂y) + ik((∂v3/∂z) = eiei(∂vi/∂xj) - - | ∂v1/∂x1 ∂v2/∂x1 ∂v3/∂x1 | = | ∂v1/∂x2 ∂v2/∂x2 ∂v3/∂x2 | | ∂v1/∂x3 ∂v2/∂x3 ∂v3/∂x3 | - - = Jacobean matrix ----------------------------------------------------- The Inner Product ----------------- Consider, the 2D case: (e1 ∧ e2).(e1 ∧ e2) (e1e2 - e2e1).(e1e2 - e2e1) (e1e2 + e1e2).(e1e2 + e1e2) 4(e1e2).(e1e2) 4(e1.e1)(e2.e2) The is the product of eigenvalues which is equivalent to the determinant of the matrix, i.e. - - = | e1.e1 0 | | 0 e2.e2 | - - = det(gij) = (-1)D-S Where S is the SIGNATURE of the metric = (# of + elements, # of - elements). If D = 2 we get gij = (1,1) ∴ S = (2,0) We use these results in the next section. The Hodge Star (*) Operator --------------------------- The Hodge Star Operator on an D-dimensional manifold is a map from p-forms to (D - p) forms. *: Λp <-> ΛD-p The meaning of 'dual' in this context does NOT mean dual as in vectors -> dual vectors. If V has an inner product defined on it then an inner product is induced on Λp(V) and vice versa. If V* has an inner product defined on it then an inner product is induced on Λp(V*) and vice versa. Schematically, metric V <---------------> V* metric Λp(V) <---------------> Λ0(V*) | | | Hodge * | Hodge * | | v v ΛD-p(V) <---------------> ΛD-p(V*) metric Consider: α, β ∈ Λp Now construct the inner product: α.β = Y Y ∈ ℝ Now take, α ∈ Λp and γ ∈ ΛD-p Write, α ∧ γ = XE α ∧ γ ∈ ΛpΛD-p = ΛD with dimension = 1 so X is a single number ∈ ℝ. E is the unit volume equal to dx0 ∧ ... dxD and also lives in ΛD so it also 1 dimensional. Because XE and Y are single numbers we can equate them. Therefore, α ∧ γ = (α.β)E When this is true, α = *β Λp ΛD-p \ / α ∧ *β = (α.β)pE \ \ Λp ΛD-p Examples: Consider *1. This is the scalar Λ0 with basis = 1. For D = 3 we have: Λ0 Λ3-0 \ / α ∧ *1 = (α.1)pE \ \ Λ0 Λ3-0 α is a scalar so α ∧ *1 = α*1. Therefore, α*1 = αE So, *1 = e1 ∧ e2 ∧ e3) Now consider *E E ∧ *E = (E.E)NE For D = 3 we have: Λ3 Λ3-3 \ / E ∧ *E = (E.E)pE \ \ Λ3 Λ3-3 E is a scalar since it has dimension = 1. So, *(e1 ∧ e2 ∧ e3) = 1 Now consider *e2: e2 ∧ *e2 = (e2.e2)pE e2 ∧ *e2 = -(e2 ∧ e1 ∧ e3) For D = 3 we have: Λ1 Λ3-1 \ / e2 ∧ *e2 = -(e2e1 ∧ e3) \ \ Λ1 Λ3-1 Compare the Λ3-1 spaces to get *e2 = -(e1 ∧ e3) As a final example consider *(e1 ∧ e3): (e1 ∧ e3) ∧ *(e1 ∧ e3) = (e1.e1)(e3.e3)(e1 ∧ e2 ∧ e3) (e1 ∧ e3) ∧ *(e1 ∧ e3) = -(e1 ∧ e3 ∧ e2) For D = 3 we have: Λ2 Λ3-2 \ / (e1 ∧ e3) ∧ *(e1 ∧ e3) = -(e1 ∧ e3 ∧ e2) \ \ Λ2 Λ3-2 Compare the Λ3-1 spaces to get *(e1 ∧ e3) = -e2 In D dimensions we can generalize these operations as follows: Let I = i1 ... ip and H = j1 ... jD-p. I spans the space Λp and H spans the space ΛD-p. Therefore, together I and H span the entire space ΛD. We can then write: *[ei1 ∧ ... ∧ eip] = εI,H[ei1.ei1 ... eip.eip]ej1 ∧ ... ∧ ejD-p Example: For D = 3 *e1 = ε1,23(e1.e1)e2 ∧ e3 = e2 ∧ e3 *e2 = ε2,13(e2.e1)e2 ∧ e3 = -(e1 ∧ e3) Geometrically, the Hodge dual captures the idea that planes can be identified with their normals and so forth. Grassmann Numbers ----------------- Grassmann numbers, θ, are anti-commuting numbers that are individual elements of the exterior algebra. Crudely speaking, they can be viewed as differential forms without the notion of a manifold or exterior deivative. Grassmann numbers are used in Physics to describe Fermionic fields where the particles have antisymmetric wavefunctions. Geometric (aka Clifford) Algebra -------------------------------- The fundamental operation in Geometric Algebra is the Geometric Product. It is defined as: uv = u.v + u ∧ v u.v = (1/2)(uv + vu) u ∧ v = (1/2)(uv - vu) Geometric algebra can be regarded as an extension of exterior algebra to include scalars and vectors. The scalars and vectors have their usual interpretation and live subspaces of the geometric algebra whereas the wedge products are multivectors that live in the exterior space, Λ(V) of a vector space, V. Axioms: (uv)w = u(vw) u(v + w) = uv + uw 2 Dimensions ------------ Consider the basis vectors e1 and e2: e1.e1 = (1/2)(e1e1 + e1e1) = 1 e1 ∧ e1 = (1/2)(e1e1 - e1e1) = 0 e1e1 = e1.e1 + e1 ∧ e1 = 1 + 0 = 1 Likewise, e2e2 = 1 e1 ∧ e2 = (1/2)(e1e2 - e2e1) = -(1/2)(e2e1 - e1e2) = -(e2 ∧ e1) e1e2 = e1.e2 + e1 ∧ e2 = e1 ∧ e2 e2e1 = e2 ∧ e1 = -(e1 ∧ e2) = -e1e2 Summary: eμeμ = 1 eμeν = -eνeμ eμ ∧ eμ = 0 eμ ∧ eν = -(eν ∧ eμ) uv = u.v + u ∧ v uv = (u1e1 + u2e2)(v1e1 + v2e2) = u1v1e1e1 + u2v2e2e2 + u1v2e1e2 + u2v1e2e1 = u1v1 + u2v2 + e1e2(u1v2 - u2v1) = u.v + e1e2(u1v2 - u2v1) Therefore, e1e2(u1v2 - u2v1) ≡ u ∧ v Proof: u ∧ v = u1v1(e1 ∧ e1) + u2v2(e2 ∧ e2) + u1v2(e1 ∧ e2) + u2v1(e2 ∧ e1) = u1v2(e1 ∧ e2) + u2v1(e2 ∧ e1) = (u1v2 - u2v1)(e1 ∧ e2) This is a bivector. Bivectors are often referred to as axial or pseudovectors. Geometrically e1e2 represent oriented areas. The algebra should be closed meaning that the product of 2 elements is another element. If we consider all possible combinations of e1 and e2 we obtain a 4-dimensional space spanned by {1,e1,e2,e1 ∧ e2}. Scalars (grade 0): 1 Vectors (grade 1): e1, e2 Bivectors (grade 2): e1 ∧ e2 In Clifford Algebra dimensions are referred to as 'grades'. Relation to Complex Numbers --------------------------- (e1e2)2 = e1e2e1e2 = -e1e1e2e2 We can therefore identify e1e2 with i since i2 = -1 We say that Cl2 is isomorphic to the complex numbers. 3 Dimensions ------------ We can extend the basis vector constructions to any number of dimensions. However, we need to be careful about how these rules are applied. For example, e1e2e1e3 = e1 ∧ e2 ∧ e1 ∧ e3 = -(e1 ∧ e1 ∧ e2 ∧ e3) = -(0 ∧ e2 ∧ e3) = 0 Is not a valid construction. To do this correctly we need to go back to the formula uv = u.v + u ∧ v and write: e1e2e1e3 = (-e1e1)(e2e3) = uv Then: uv = (-e1e1).(e2e3) + (-e1e1) ∧ (e2.e3) = -1.(e2e3) + -1 ∧ e2 ∧ e3 = 0 - (e2 ∧ e3) = -(e2 ∧ e3) Of course, the shorter calculation is just to simplify the original expression before constructing the wedge product. Thus, e1e2e1e3 = -e1e1e2e3 = -e2e3 = -(e2 ∧ e3) Again, we look for the closure requirement by considering at all possible combinations of e1, e2 and e3: (e1e2)e1 = -e2 (e1e2)e2 = e1 (e1e2)e3 = e1e2e3 (e2e3)e1 = e2e3e1 = e1e2e3) (e2e3)e2 = -e3 (e2e3)e3 = e2 (e1e3)e1 = -e3 (e1e3)e2 = e1e3e2 = -(e1e2e3) (e1e3)e3 = e1 Therefore we obtain an 8-dimensional space spanned by {1,e1,e2,e3,e1 ∧ e2,e2 ∧ e3,e1 ∧ e3,e1 ∧ e2 ∧ e3} Scalars (grade 0): 1 Vectors (grade 1): e1, e2, e3 Bivectors (grade 2): e1 ∧ e2, e2 ∧ e3, e1 ∧ e3 Trivectors (grade 3): e1 ∧ e2 ∧ e3 Trivectors are often referred to as pseudoscalars. Geometrically, pseudoscalars e1e2e3 represent oriented volumes. It is worth noting that the bases of the algebra follows Pascal's triangle. 0: 1 1: 1 1 2: 1 2 1 3: 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 Which is tantamount to saying that the dimension of the Clifford algebra generated by a vector space of dimension, D, is given by 2D. Relation to Quaternions ----------------------- i2 = j2 = k2 = ijk = -1 i2 = e1e2e1e2 = -e1e1e2e2 = -1 j2 = e2e3e2e3 = -e2e2e3e3 = -1 k2 = e1e3e1e3 = -e1e1e3e3 = -1 ijk = e1e2e2e3e1e3 = e1e3e1e3 = -e1e1e3e3 = -1 We say that Cl3 is isomorphic to the quaternions. Examples in 3D -------------- Vector: u = (3e2 + 4e3) Bivector: v = (2e1e2 + 3e2e3 + 4e1e3) uv = (3e2 + 4e3)(2e1e2 + 3e2e3 + 4e1e3) = 6e2e1e2 + 9e2e2e3 + 12e2e1e3 + 8e3e1e2 + 12e3e2e3 + 16e3e1e3 = -6e1 + 9e3 - 12e1e2e3 + 8e1e2e3 - 12e2 - 16e1 = -22e1 - 12e2 + 9e3 - 4e1e2e3 Compare this result to the Grassman algebra where cannot simplify the basis vectors, e1 ... : u ∧ v = (3e2 + 4e3) ∧ (2e1e2 + 3e2e3 + 4e1e3) = (3e2 ∧ 2e1e2) + (3e2 ∧ 3e2e3) + (3e2 ∧ 4e1e3) + (4e3 ∧ 2e1e2) + (4e3 ∧ 3e2e3) + (4e3 ∧ 4e1e3) = (3 ∧ e2 ∧ 2 ∧ e1 ∧ e2) + (3 ∧ e2 ∧ 3 ∧ e2 ∧ e3) + (3 ∧ e2 ∧ 4 ∧ e1 ∧ e3) + (4 ∧ e3 ∧ 2 ∧ e1 ∧ e2) + (4 ∧ e3 ∧ 3 ∧ e2 ∧ e3) + (4 ∧ e3 ∧ 4 ∧ e1 ∧ e3) = 12(e2 ∧ e1 ∧ e3) + 4(e3 ∧ 2e1 ∧ e2) = -12(e1 ∧ e2 ∧ e3) + 8(e1 ∧ e2 ∧ e3) = -4(e1 ∧ e2 ∧ e3) Bivector: u = (5 + 2e1e2) Bivector: v = (2 + 2e1 + e1e3) uv = (5 + 2e1e2)(2 + 2e1 + e1e3) = 10 + 10e1 + 5e1e3 + 4e1e2 + 4e1e2e1 + 2e1e2e1e3 = 10 + 10e1 - 4e2 + 5e1e3 + 4e1e2 - 2e2e3 -- ---------- -------------------- ^ ^ ^ | | | | vector bivectors scalar We see that the geometric product of a bivector with a bivector does not expand to a quadrivector, but rather it reduces to a bivector. This 'wrapping' is a form of closure in that runaway expansion into ever higher dimensions is prevented. We can see this in 3D as follows: (vector)(vector) = (e1)(e2) = e1e2 (bivector) (vector)(bivector) = (e1)(e2e3) = e1e2e3 (trivector) (vector)(trivector) = (e1)(e1e2e3) = e2e3 (bivector) (bivector)(bivector) = (e1e2)(e1e3) = -e2e3 (bivector) (bivector)(trivector) = (e1e2)(e1e2e3) = -e3 (vector) Bivector: u = (5 + 2e1e2) Trivector: v = (2 + e1e2e3) uv = (5 + 2e1e2)(2 + e1e2e3) = 10 + 5e1e2e3 + 4e1e2 + 2e1e2e1e2e3 = 10 + 5e1e2e3 + 4e1e2 - 2e3 Now, for an orthogonal basis e1e2 = e1 ∧ e2 etc. uv = 10 + 5(e1 ∧ e2 ∧ e3) + 4(e1 ∧ e2) - 2e3 = 10 - 2e3 + 4(e1 ∧ e2) + 5(e1 ∧ e2 ∧ e3) -- --- ---------- --------------- ^ ^ ^ ^ | | | | | vector bivector trivector scalar Matrix Representation --------------------- Consider the Pauli matrices. σ1σ2 - σ2σ1 = 2iσ3 - - - - - - - - - - | 0 1 || 0 -i | - | 0 -i || 0 1 | = | 2i 0 | | 1 0 || i 0 | | i 0 || 1 0 | | 0 -2i | - - - - - - - - - - If we let σ1 ≡ e1, σ2 ≡ e2 and σ3 ≡ e3 (1/2)(e1e2 - e2e1) = e1 ∧ e2 ≡ e1e2e3e3 ≡ Ie3 Where I = e1e2e3 Or, (e1e2 - e2e1) = 2Ie3 The Clifford Algebra for the Pauli matrices is often written as: {σμν} = 2ημνI Which yields: σμσν + σνσμ = 2ημνI μ = ν => σ12 = σ22 = σ32 = 1 μ ≠ ν => σ1σ2 + σ2σ1 = 2η12I = 0 ∴ σ1σ2 = -σ2σ1 It should not be surprising that the Dirac matrices also also obey the Clifford algebra. {γμν} = 2ημνI